Answer
$\{-3,0,3\}$
Work Step by Step
We are given the trinomial:
$$p(x)=4x^2+bx-1.$$
As the trinomial is a quadratic function, it can be factored if it has real root(s)
$$\begin{align*}
b^2-4(4)(-1)&\geq 0\\
b^2+16&\geq 0.
\end{align*}$$
So the trinomial can be factored for any real value of $b$.
$$4x^2+bx-1=(4x+u)(x+v)\text { or } 4x^2+bx-1=(2x+u)(2x+v).$$
$\textbf{Case 1:}$ $4x^2+bx-1=(4x+u)(x+v)$
$$\begin{align*}
4x^2+bx-1&=4x^2+(4v+u)x+uv\\
b&=4v+u\\
uv&=-1\\
u=1,v=-1&\text{ or }u=-1,v=1\\
b=4(-1)+1\text{ or }b=4(1)+(-1)\\
b&=3&\text{ or }b=-3.
\end{align*}$$
$\textbf{Case 2:}$ $4x^2+bx-1=(2x+u)(2x+v)$
$$\begin{align*}
4x^2+bx-1&=4x^2+(2u+2v)x+uv\\
b&=2u+2v\\
uv&=-1\\
u=1,v=-1&\text{ or }u=-1,v=1\\
b=2(1)+2(-1)\text{ or }b=2(-1)+2(1)\\
b&=0.
\end{align*}$$
So the integer values of $b$ are $\{-3,0,3\}$.
